This preview shows page 1. Sign up to view the full content.
46. The
q
in the denominator is to be interpreted as

q

(so that the orbital radius
r
is a positive number).
We interpret the given 10
.
0 MeV to be the kinetic energy of the electron. In order to make use of the
mc
2
value for the electron given in Table 383 (511 keV = 0
.
511 MeV) we write the classical kinetic energy
formula as
K
classical
=
1
2
mv
2
=
1
2
(
mc
2
)
µ
v
2
c
2
¶
=
1
2
(
mc
2
)
β
2
.
(a) If
K
classical
=10
.
0MeV, then
β
=
r
2
K
classical
mc
2
=
r
2(10
.
0MeV)
0
.
511 MeV
=6
.
256
,
which, of course, is impossible (see the Ultimate Speed subsection of
§
382). If we use this value
anyway, then the classical orbital radius formula yields
r
=
mv

q

B
=
mβc
eB
=
(
9
.
11
×
10
−
31
kg
)
(6
.
256)
(
2
.
998
×
10
8
m
/
s
)
(1
.
6
×
10
−
19
C) (2
.
20 T)
=4
.
85
×
10
−
3
m
.
If, however, we use the correct value for
β
(calculated in the next part) then the classical radius
formula would give about 0
.
77 mm.
(b) Before using the relativistically correct orbital radius formula, we must compute
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy, Kinetic Energy

Click to edit the document details